Ngā Tauira Pātai e Matapaki ana i te Kaha i runga i te Utu Neke
Pendahuluan
Ko te ahupūngao te ako i ngā āhuatanga taiao, tae atu ki ngā kaha e pā ana ki ngā mea. Ko tētahi kaupapa whakamere e kōrerohia pinepinetia ana ko te kaha ki ngā utu neke, inā koa i roto i te horopaki o ngā papa hiko me ngā papa aukume. Ko te kaha e pā ana ki tētahi utu neke i roto i tētahi papa hiko, papa aukume rānei, e mōhiotia ana ko te kaha Lorentz. Ka matapakihia e tēnei tuhinga ētahi tauira raruraru me ā rātou matapakinga e pā ana ki te kaha e pā ana ki ngā utu neke.
Te kaha o Lorentz
Ko te kaha Lorentz he huinga o te kaha hiko me te kaha aukume e mahi ana i runga i te utu e neke ana i roto i te papa hiko me te papa aukume. Mā te pāngarau, ka taea te whakaatu i te kaha Lorentz (F) mā te whārite:
\[ \mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \]
Kei hea:
– \( \mathbf{F} \) ko te kaha Lorentz
– Ko te utu te \( q \)
– Ko te mara hiko te \( \mathbf{E} \)
– Ko te tere o te utunga \( \mathbf{v} \)
– Ko te papa aukume te \( \mathbf{B} \)
Mā tēnei whārite, ka taea e tātou te tātari i ngā kaha e pā ana ki tētahi utu e neke ana i roto i tētahi papa hiko me tētahi papa aukume.
Ngā Pātai Tauira me te Kōrero
Pātai 1: Te kaha i runga i te utu i roto i te mara hiko
Pātai:
Kei roto i tētahi papa hiko ōrite te utu pai \( q = 2 \times 10^{-6} \, C \) \( E = 5 \times 10^4 \, N/C \) e anga ana ki te taha matau. Tātaihia te kaha e pā ana ki te utu.
Kōrero:
Mō te utu i roto i te papa hiko, kāore he papa aukume, ko te kaha Lorentz he mea hanga noa iho e te wāhanga kaha hiko:
\[ \mathbf{F} = q \mathbf{E} \]
Me \( q = 2 \times 10^{-6} \, C \) me \( \mathbf{E} = 5 \times 10^4 \, N/C \):
\[ \mathbf{F} = (2 \times 10^{-6} \, C) \times (5 \times 10^4 \, N/C) \]
\[ \mathbf{F} = 0.1 \, N \]
Ko te ahunga o te kaha \( \mathbf{F} \) kei te ahunga kotahi ki te mara hiko nā te mea he pai te utu. Nō reira, ko te kaha e pā ana ki te utu he 0.1 N ki te taha matau.
Pātai 2: Te kaha i runga i te utu i roto i te papa aukume
Pātai:
Ka neke te utunga kino \( q = -3 \times 10^{-6} \, C \) me te tere \( \mathbf{v} = 2 \times 10^3 \, m/s \) i te tuaka-x i roto i te papa aukume ōrite \( \mathbf{B} = 0.5 \, T \) e anga ana i te tuaka-z. Tātaihia te kaha e pā ana ki te utunga.
Kōrero:
Mō te utu e neke ana i roto i te papa aukume me te kore he papa hiko, ko te kaha Lorentz he mea hanga noa iho i ngā wāhanga o te kaha aukume:
\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]
Me te \( q = -3 \times 10^{-6} \, C \), \( \mathbf{v} = 2 \times 10^3 \, m/s \) i te ahunga x, me te \( \mathbf{B} = 0.5 \, T \) i te ahunga z:
Tātaitanga \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = 2 \times 10^3 \, m/s \, \hat{i} \]
\[ \mathbf{B} = 0.5 \, T \, \potae{k} \]
\[ \mathbf{v} \times \mathbf{B} = (2 \times 10^3 \, m/s \, \hat{i}) \times (0.5 \, T \, \hat{k}) \]
\[ \mathbf{v} \times \mathbf{B} = 2 \times 10^3 \, m/s \times 0.5 \, T \, \hat{i} \times \hat{k} \]
\[ \hat{i} \times \hat{k} = -\hat{j} \]
\[ \mathbf{v} \times \mathbf{B} = – (1 \times 10^3 \, T \cdot m/s) \, \hat{j} \]
Nā, ko te kaha aukume:
\[ \mathbf{F} = q \mathbf{v} \times \mathbf{B} \]
\[ \mathbf{F} = (-3 \times 10^{-6} C) \times ( -10^3 \, T \cdot m/s \, \hat{j}) \]
\[ \mathbf{F} = 3 \wā 10^{-3} \, N \, \pōtae{j} \]
\[ \mathbf{F} = 0.003 \, N \, \potae{j} \]
Ko te ahunga o te kaha \( \mathbf{F} \) e anga ana ki te tuaka-y pai. Nō reira, ko te kaha e pā ana ki te utu he 0.003 N ki runga (i te ahunga o te tuaka-y pai).
Pātai 3: Te Kaha ki te Utu i roto i ngā Papa Hiko me ngā Papa Aukume
Pātai:
Ka neke te utu pai \( q = 1.5 \times 10^{-6} \, C \) me te tere \( \mathbf{v} = 4 \times 10^3 \, m/s \) i te ahunga-y i roto i te papa hiko \( \mathbf{E} = 3 \times 10^4 \, N/C \) i te ahunga-x, me te papa aukume \( \mathbf{B} = 0.2 \, T \) i te ahunga-z. Tātaihia te kaha kupenga e pā ana ki te utu.
Kōrero:
Te tapeke o te kaha Lorentz:
\[ \mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) \]
Tuatahi, tatauhia \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = 4 \times 10^3 \, m/s \, \hat{j} \]
\[ \mathbf{B} = 0.2 \, T \, \potae{k} \]
\[ \mathbf{v} \times \mathbf{B} = (4 \times 10^3 \, m/s \, \hat{j}) \times (0.2 \, T \, \hat{k}) \]
\[ \mathbf{v} \times \mathbf{B} = 4 \times 10^3 \, m/s \times 0.2 \, T \, \hat{j} \times \hat{k} \]
\[ \hat{j} \times \hat{k} = \hat{i} \]
\[ \mathbf{v} \times \mathbf{B} = (0.8 \times 10^3 \, T \cdot m/s) \, \hat{i} \]
\[ \mathbf{v} \times \mathbf{B} = 800 \, T \cdot m/s \, \hat{i} \]
Kātahi, ko te kaha hiko:
\[ q \mathbf{E} = (1.5 \times 10^{-6} \, C) \times (3 \times 10^4 \, N/C \, \hat{i}) \]
\[ q \mathbf{E} = 0.045 \, N \, \hat{i} \]
Te kaha aukume:
\[ q (\mathbf{v} \times \mathbf{B}) = (1.5 \times 10^{-6} \, C) \times (800 \, T \cdot m/s \, \hat{i}) \]
\[ q (\mathbf{v} \times \mathbf{B}) = 0.0012 \, N \, \hat{i} \]
Kāhua katoa:
\[ \mathbf{F} = q \mathbf{E} + q (\mathbf{v} \times \mathbf{B}) \]
\[ \mathbf{F} = 0.045 \, N \, \pōtae{i} + 0.0012 \, N \, \ pōtae {i} \]
\[ \mathbf{F} = 0.0462 \, N \, \ pōtae {i} \]
Nō reira, ko te kaha katoa e pā ana ki te utu he 0.0462 N ki te taha matau (tuaka-x pai).
Whakamutunga
Ko te kaha o ngā utu e neke ana i roto i ngā papa hiko me ngā papa aukume e whakawhirinaki nui ana ki te ahunga me te rahi o ia papa me te tere me te momo utu. Mā roto i ngā tauira pātai me te matapaki i runga ake nei, ko te tumanako ka pai ake te mārama o ngā kaipānui ki te whakamahi i te mātāpono kaha Lorentz i roto i ngā āhuatanga rerekē. Ehara i te mea he mea nui tēnei māramatanga i roto i te ariā anake engari i roto hoki i ngā tono mahi i roto i ngā mara hangarau me te pūtaiao pērā i te hoahoa o ngā motuka hiko, te mārama ki te āhuatanga aurora, me te mahi a ngā matūriki i roto i ngā whakatere matūriki.
